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Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

225 Q 5 Topics Take Test

Difficulty: All Easy Medium Hard 11–20 of 225

Topics in JEE Chemistry

All Physical Chemistry 100 Organic Chemistry 100 Inorganic Chemistry 100 Electrochemistry 100 Chemical Kinetics 57

Q.11 Medium Chemical Kinetics

The temperature coefficient (Q₁₀) for a reaction is 2.5. If the rate at 300 K is r, then the rate at 320 K is approximately:

A 2.5r

B 6.25r

C 5r

D 12.5r

Correct Answer: B. 6.25r

EXPLANATION

Q₁₀ = rate at (T+10)/rate at T. For 300K to 320K (two 10K intervals), rate = r × 2.5² = 6.25r

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Q.12 Medium Chemical Kinetics

In enzyme catalysis, the Michaelis constant (Km) represents:

A Maximum enzyme velocity

B The substrate concentration at half maximum velocity

C The enzyme concentration

D The activation energy of the reaction

Correct Answer: B. The substrate concentration at half maximum velocity

EXPLANATION

Km is a characteristic constant for an enzyme-substrate pair, representing substrate concentration when v = Vmax/2

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Q.13 Medium Chemical Kinetics

The reaction A → B has rate constant k = 4.5 × 10⁻³ min⁻¹. This is a first-order reaction. The percentage of A remaining after 100 minutes is:

A 23.5%

B 35.7%

C 41.2%

D 58.8%

Correct Answer: C. 41.2%

EXPLANATION

[A]ₜ = [A]₀e^(-kt) = 100 × e^(-0.0045 × 100) = 100 × e^(-0.45) ≈ 41.2%

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Q.14 Medium Chemical Kinetics

For a bimolecular reaction between A and B molecules, the collision frequency (Z) depends on:

A Only temperature

B Only molecular size and concentration

C Only collision cross-section

D Temperature, molecular size, concentration, and average relative velocity

Correct Answer: D. Temperature, molecular size, concentration, and average relative velocity

EXPLANATION

Z ∝ σ × [A] × [B] × √(T/M), where all these factors contribute to collision frequency

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Q.15 Medium Chemical Kinetics

In the reaction 2A + B → C, if the concentration of A is doubled and B is tripled, the rate increases by 12 times. The rate law is:

A Rate = k[A][B]

B Rate = k[A]²[B]

C Rate = k[A][B]²

D Rate = k[A]²[B]²

Correct Answer: C. Rate = k[A][B]²

EXPLANATION

If Rate = k[A]^m[B]^n, then 12 = 2^m × 3^n. Testing: 2¹ × 3² = 2 × 9 = 18 (no); 2² × 3¹ = 12 (yes). So m=1, n=2, giving Rate = k[A][B]²

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Q.16 Medium Chemical Kinetics

The half-life of a first-order reaction is independent of the initial concentration. If t₁/₂ = 30 minutes for a reaction, the time for the concentration to reduce to 1/4th of initial value is:

A 30 minutes

B 45 minutes

C 60 minutes

D 90 minutes

Correct Answer: C. 60 minutes

EXPLANATION

For first-order reaction, [A]ₜ = [A]₀(1/2)^(t/t₁/₂). For [A]ₜ = 1/4[A]₀, we need (1/2)^(t/30) = 1/4, so t/30 = 2, giving t = 60 minutes

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Q.17 Medium Chemical Kinetics

In the decomposition of N₂O₅, the rate constant at 320 K is 1.7 × 10⁻⁵ s⁻¹ and at 330 K is 5.0 × 10⁻⁵ s⁻¹. The activation energy is approximately:

A 50 kJ/mol

B 100 kJ/mol

C 150 kJ/mol

D 200 kJ/mol

Correct Answer: A. 50 kJ/mol

EXPLANATION

Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂): ln(5.0/1.7) = (Ea/8.314)(1/320 - 1/330), solving gives Ea ≈ 50 kJ/mol

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Q.18 Medium Chemical Kinetics

A reaction has activation energy of 50 kJ/mol. If the temperature is increased from 300 K to 310 K, the rate constant increases by a factor of approximately (R = 8.314 J/mol·K):

A 1.5

B 2.0

C 2.5

D 3.0

Correct Answer: B. 2.0

EXPLANATION

Using Arrhenius equation: log(k₂/k₁) = (Ea/2.303R)(T₂-T₁)/(T₁T₂). With Ea = 50,000 J/mol, ΔT = 10 K, this gives log(k₂/k₁) ≈ 0.30, so k₂/k₁ ≈ 2.0

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Q.19 Medium Chemical Kinetics

Which of the following is an example of a homogeneous catalyst?

A Fe₃O₄ in ammonia synthesis

B H₂SO₄ in esterification reaction

C Pt surface in hydrogenation

D MnO₂ in decomposition of KMnO₄

Correct Answer: B. H₂SO₄ in esterification reaction

EXPLANATION

A homogeneous catalyst is in the same phase as reactants. H₂SO₄ (liquid) catalyzes esterification of reactants (liquid), making it homogeneous. Others are heterogeneous catalysts.

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Q.20 Medium Chemical Kinetics

If a reaction is first-order with rate constant k = 0.1 min⁻¹, what fraction of the reactant remains after 5 half-lives?

A 1/32

B 1/16

C 1/8

D 1/4

Correct Answer: A. 1/32

EXPLANATION

After n half-lives, fraction remaining = (1/2)ⁿ. After 5 half-lives: (1/2)⁵ = 1/32.

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